17.4Rate Laws
Section · module m68789
What this needs
Learn these first — this section will not make sense without them.
Solving 2.25 = 1.5^m for the reaction order m requires the log-of-a-power rule ln(1.5^m) = m·ln(1.5) to pull the exponent down before dividing; without that rule the exponent stays trapped inside the power and the method-of-initial-rates worked example cannot be finished algebraically.
The whole method of initial rates, once ratios stop being clean integers, reduces to solving aᵐ = b for the unknown exponent m by taking logs of both sides — the exponential-equation technique itself; eyeballing that doubling [NO] doubles the rate works for m = 1 but breaks the moment a ratio like 2.25 or a fractional order appears.
What this touches
Algebra that turns up here. Not a blocker, but this is where you will see it used.
Writing [CO]⁰ = 1 to drop a reactant from a rate law, and tracking rate-constant units like L² mol⁻² s⁻¹ through a calculation, both lean on the zero-exponent and negative-exponent rules a student already has from scientific notation — chemistry is simply where those exponent rules get exercised on messy mixed units instead of clean powers of ten.